Concave and convex lenses are placed touching each other. The ratio of magnitudes of their power is \(2: 3\). The focal length of the system is 30 cm . The focal lengths of individual lens are

Power \((P)=\frac{1}{f}...(i)\)
Given, \(\frac{\mathrm{P}_{\text {concave }}}{\mathrm{P}_{\text {convex }}}=\frac{2}{3}...(ii)\)
Let the focal length of the convex lens be \(\mathrm{f}_{\text {convex }}=\mathrm{f}\)
From (i) and (ii),
\(\mathrm{f}_{\text {concave }}=-\frac{3}{2} \mathrm{f} \cdot(-\mathrm{ve}\) as lens is concave \()\)
Using the equivalent focal length for two lenses in contact, we get
\(\frac{1}{f_{e q}} =\frac{1}{f}+\frac{1}{-\frac{3}{2} f} \)
\( \Rightarrow \frac{1}{30} =\frac{1}{f}-\frac{2}{3 f} \)
\( =\frac{3 f-2 f}{3 f^2}\)
\(\Rightarrow \frac{1}{30}=\frac{1}{3 \mathrm{f}} \)
\( \therefore \quad, \mathrm{f}=10 \mathrm{~cm} \)
\( \therefore \mathrm{f}_{\text {convex }}=10 \mathrm{~cm} \text { and } \mathrm{f}_{\text {concave }}=\frac{-3}{2} \times 10\) \(=-15 \mathrm{~cm}\)